Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;

vector<int> bfs(vector<vector<int>> &G, int s){
    vector<int> d(G.size(), MOD);
    d[s] = 0;
    queue<int> Q;
    Q.emplace(s);
    while(!Q.empty()){
        int i = Q.front(); Q.pop();
        for (auto &&j : G[i]) {
            if(d[i] + 1 < d[j]){
                d[j] = d[i] + 1;
                Q.emplace(j);
            }
        }
    }
    return d;
}

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    s--; t--;
    vector<vector<int>> G(n);
    for (int i = 0; i < m; ++i) {
        int x, y;
        scanf("%d %d", &x, &y);
        x--; y--;
        G[x].emplace_back(y);
        G[y].emplace_back(x);
    }
    auto d1 = bfs(G, s), d2 = bfs(G, t);
    vector<int> cnt(n);
    for (auto &&j : d2) {
        if(j < n)cnt[j]++;
    }
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
        int to = d1[t]-d1[i]-2;
        if(to >= 0 && to < n){
            ans += cnt[to];
        }
    }
    cout << ans << "\n";
    return 0;
}